A hydraulic automobile lift is designed to lift cars with a maximum mass of `3000Kg`. The area of cross section of the piston carrying the load is `425cm^(2)`. What maximum pressures would the smaller piston have to bear?

Explanation

To solve the problem of determining the maximum pressure that the smaller piston in a hydraulic lift would have to bear, we can follow these steps:

### Step-by-Step Solution:

**Step 1: Identify the mass of the car.**
- Given mass (m) = 3000 kg.

**Step 2: Calculate the weight of the car.**
- The weight (W) can be calculated using the formula:
\[ W = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
- Thus,
\[ W = 3000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 29400 \, \text{N} \]

**Step 3: Convert the area of the piston from cm² to m².**
- Given area (A) = 425 cm².
- To convert to m², use the conversion factor \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \):
\[ A = 425 \, \text{cm}^2 \cdot 10^{-4} \, \text{m}^2/\text{cm}^2 = 0.0425 \, \text{m}^2 \]

**Step 4: Calculate the pressure exerted on the piston.**
- Pressure (P) is defined as force (F) divided by area (A):
\[ P = \frac{F}{A} \]
- Here, \( F \) is the weight of the car (29400 N) and \( A \) is the area of the piston (0.0425 m²):
\[ P = \frac{29400 \, \text{N}}{0.0425 \, \text{m}^2} \approx 691764.71 \, \text{N/m}^2 \]

**Step 5: Convert the pressure to standard scientific notation.**
- The pressure can be expressed in scientific notation:
\[ P \approx 6.92 \times 10^5 \, \text{N/m}^2 \]

### Final Answer:
The maximum pressure that the smaller piston would have to bear is approximately \( 6.92 \times 10^5 \, \text{N/m}^2 \).

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